The provided image shows a mathematical derivation for an infinite series. Let's break down the work presented:

1. Original Series Setup: The problem begins with an infinite series: $\sum_{n=3}^\infty \frac{5 + 7n - 3n^2}{[n(n+1)]^2}.$

2. Substitution of Variables: A substitution is made, setting $n = k+1$, which shifts the series starting index from $n=3$ to $k=2$. Under this transformation: $n = k+1 \implies 3 = k+1 \implies k = 2.$ After substitution, the series becomes: $\sum_{k=2}^\infty \frac{5 + 7(k+1) - 3(k+1)^2}{[(k+1)k(k+2)]^2}.$

3. Simplification: The numerator is simplified as: $5 + 7(k+1) - 3(k+1)^2 = 5 + 7k + 7 - 3(k^2 + 2k + 1) = 9 + k - 3k^2.$ The series becomes: $\sum_{k=2}^\infty \frac{9 + k - 3k^2}{[(k+1)(k+2)]^2}.$

4. Decomposition: For $k=2$, the first term is calculated separately: [ \frac{9 + 2 - 3(2)^2}{[(2+1)(2+2)]^2} = \frac{-7}{[3 \cdot 4]^2} = \frac{-7}{144}. ]

   The remaining series continues for $k \geq 3$: $\sum_{k=3}^\infty \frac{9 + k - 3k^2}{[(k+1)(k+2)]^2}.$

5. Final Summation: The terms are combined and simplified. The result of the sum is: $-\frac{4}{9}.$

Would you like a detailed explanation of any particular step or have questions about the technique?

Here are five relative questions to expand understanding:

1. What is the rationale behind using substitutions like $n = k+1$ in summations?
2. How can the numerator simplification $9 + k - 3k^2$ be verified step-by-step?
3. What alternative methods exist to compute or approximate infinite series like this one?
4. How does the initial $k=2$ term impact the convergence of the series?
5. Can this technique be applied to summations involving factorials or exponential terms?

Tip: Breaking a complex series into parts (like isolating the first few terms) often